By P. Sabine, C. Plumpton (auth.)

ISBN-10: 0333383648

ISBN-13: 9780333383643

ISBN-10: 1349076686

ISBN-13: 9781349076680

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P( = t < - ~:~~~ x yI(25)) = pet < -2) = 1 - <1>(2) = 0·0228. Example 6 Television tubes produced by a certain company are known to have a mean lifetime of 90 months with standard deviation 8 months . Fifty samples of 100 tubes produced by this company are tested. Estimate, to 2 decimal places, the number of samples which would be expected to have a mean lifetime of more than 91 months. Let X months be the lifetime of a tube. Then X is distributed with mean 90 months and standard deviation 8 months.

Then T = N(26, 52). Here we have a single reading of 15 from the T population . As suming that this reading does come from the T population (thi s is H o), we find that peT ~ 15) = <1>(-11/5) = 1 - <1>(2·2) = 0·0139. Our test is onl y on e-tailed since we are concerned onl y with T being as lo w as or lo wer than 15. We will use a 5% significa nce level. There is onl y a probability of 0·0139 (approxim at ely 1·4 % ) of picking a car with a 'first major breakdown time' as low as or lower than this by random sampling from the Significance testing 45 population of cars being produced.

We had i = 0·287 seconds. L" (x, - i)2 for the ten reaction times is equal r=l to [0,0313 + 0·0279 + 0·0114 + 0·0161 + 0·0022 + 0·2052 + 0·0204 + 0·0543 + 0·0350 + 0,0003] (seconds)? = 0·4041 (seconds)". Also n = 10. Therefore , 0"2 = 0·4041 (10 _ 1) (seconds)? = 0·045 (seconds)". e . an estimate of the population mean {L = 0·287 seconds in the first example, and an estimate of the population variance &2 = 0·045 (seconds)? in the second example, are called point estimates of the parameter concerned .

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Statistics by P. Sabine, C. Plumpton (auth.)

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